Question
Download Solution PDFThe network using CSMA/CD has a bandwidth of 20 mbps. If the maximum propagation time is 25μ sec (microsecond), what is the minimum size of the frame?
Answer (Detailed Solution Below)
Option 2 : 1000 bits
Detailed Solution
Download Solution PDFThe correct answer is 2) 1000 bits.
Key Points
- The network uses CSMA/CD, which stands for Carrier Sense Multiple Access with Collision Detection.
- To determine the minimum frame size in a CSMA/CD network, we need to ensure that the transmission time is at least twice the maximum propagation delay. This is to detect collisions before the entire frame is transmitted.
- Given:
- Bandwidth (B) = 20 Mbps
- Maximum Propagation Time (T) = 25 μs
- The minimum frame size (S) can be calculated using the formula: S = 2 * B * T
- Calculating the minimum frame size:
- S = 2 * 20 Mbps * 25 μs
- Converting units: 1 Mbps = 10^6 bits per second and 1 μs = 10^-6 seconds
- S = 2 * 20 * 10^6 bits/second * 25 * 10^-6 seconds
- S = 2 * 20 * 25 bits
- S = 1000 bits
- However, the actual minimum frame size to avoid collision detection issues is often standardized, and in this case, it is 1000 bits.
Additional Information
- CSMA/CD is used in Ethernet networks to manage the transmission of data and avoid collisions. The calculation ensures that the round-trip time for a signal to propagate across the network and back is taken into account.
- Understanding frame size calculations is crucial for network design and ensuring efficient data transmission.